A) \[\sqrt{18}\] cm
B) \[\sqrt{20}\] cm
C) 8 cm
D) 6 cm
Correct Answer: D
Solution :
(d):- In \[\Delta ABC\] and \[\Delta ACD\], \[\therefore \] \[\frac{AC}{AB}=\frac{4}{AC}\] \[\therefore \] \[A{{C}^{2}}=4\times 13=52\] In \[\Delta ABC\] and \[\Delta BCD\], \[\frac{BC}{AB}=\frac{9}{BC}\Rightarrow B{{C}^{2}}=9\times 13=117\] Now, \[=\frac{1}{C{{D}^{2}}}=\frac{1}{A{{C}^{2}}}+\frac{1}{B{{C}^{2}}}=\frac{1}{52}+\frac{1}{117}\] \[=\frac{9+4}{13\times 4\times 9}\Rightarrow \frac{1}{C{{D}^{2}}}=\frac{1}{36}\] \[\therefore \] \[CD=6\]cm
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